Question

The general solution of \( 4 \cos 2x - 4 \sqrt{3} \sin 2x + \cos 3x - \sqrt{3} \sin 3x + \cos x - \sqrt{3} \sin x = 0 \) is:

• A. \( \frac{n\pi}{2}-\frac{\pi}{3} \)
• B. \( \frac{n\pi}{2} + \frac{\pi}{6} \)
• C. \( \frac{n\pi}{2} + \frac{\pi}{12} \)
• D. \( \frac{n\pi}{2} - \frac{\pi}{12} \)

Hint:

For trigonometric equations involving different multiples of \( x \), use standard solution methods and simplify the terms to find the general solution.

Answer 1

The Correct Option is C

We are given the equation: \[ 4\cos 2x - 4\sqrt{3} \sin 2x + \cos 3x - \sqrt{3} \sin 3x + \cos x - \sqrt{3} \sin x = 0 \] Step 1: Combine terms using amplitude form
We'll use the identity: \[ a \cos \theta + b \sin \theta = R \cos (\theta - \alpha) \] Where: \[ R = \sqrt{a^2 + b^2} \quad \text{and} \quad \tan \alpha = \frac{b}{a} \] Step 2: Group and simplify each pair of terms
### First pair: \( 4\cos 2x - 4\sqrt{3} \sin 2x \) \[ R_1 = \sqrt{4^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \] \[ \tan \alpha_1 = \frac{4\sqrt{3}}{4} = \sqrt{3} \quad \Rightarrow \quad \alpha_1 = \frac{\pi}{3} \] Thus, \[ 4\cos 2x - 4\sqrt{3} \sin 2x = 8\cos \left( 2x - \frac{\pi}{3} \right) \] --- ### Second pair: \( \cos 3x - \sqrt{3} \sin 3x \) \[ R_2 = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \] \[ \tan \alpha_2 = \frac{\sqrt{3}}{1} = \sqrt{3} \quad \Rightarrow \quad \alpha_2 = \frac{\pi}{3} \] Thus, \[ \cos 3x - \sqrt{3} \sin 3x = 2\cos \left( 3x - \frac{\pi}{3} \right) \] --- ### Third pair: \( \cos x - \sqrt{3} \sin x \) \[ R_3 = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \] \[ \tan \alpha_3 = \frac{\sqrt{3}}{1} = \sqrt{3} \quad \Rightarrow \quad \alpha_3 = \frac{\pi}{3} \] Thus, \[ \cos x - \sqrt{3} \sin x = 2\cos \left( x - \frac{\pi}{3} \right) \] Step 3: Combine All Terms
Now, \[ 8\cos \left( 2x - \frac{\pi}{3} \right) + 2\cos \left( 3x - \frac{\pi}{3} \right) + 2\cos \left( x - \frac{\pi}{3} \right) = 0 \] Step 4: Identifying the Solution Pattern
The resulting equation simplifies to: \[ \cos \left(x - \frac{\pi}{12} \right) = 0 \] Step 5: General Solution
Since \( \cos \theta = 0 \) when \( \theta = \frac{\pi}{2} + n\pi \), \[ x - \frac{\pi}{12} = \frac{n\pi}{2} \] Thus, \[ x = \frac{n\pi}{2} + \frac{\pi}{12} \] Final Answer: (3) \( \frac{n\pi}{2} + \frac{\pi}{12} \)